試證:不論正數(shù)a,b,c是等差數(shù)列還是等比數(shù)列,當n>1,n∈N且a,b,c互不相等時,都有an+cn>2bn.(n∈N).
【答案】
分析:首先題目要求證明不等式對等比數(shù)列或等差數(shù)列均成立,考慮到用數(shù)學歸納法證明,本題中使用到結論有 (a
k-c
k)(a-c)>0恒成立(a、b、c為正數(shù)),從而a
k+1+c
k+1>a
k•c+c
k•a.即可得到答案.
解答:證明 (1)設a、b、c為等比數(shù)列,a=
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,c=bq(q>0且q≠1)
∴a
n+c
n=
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+b
nq
n=b
n(
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+q
n)>2b
n
(2)設a、b、c為等差數(shù)列,
則2b=a+c猜想
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>
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(n≥2且n∈N
*)
下面用數(shù)學歸納法證明
①當n=2時,由2(a
2+c
2)>(a+c)
2,∴
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②設n=k時成立,即
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.
則當n=k+1時,
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(a
k+1+c
k+1+a
k+1+c
k+1)>
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(a
k+1+c
k+1+a
k•c+c
k•a)=
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(a
k+c
k)(a+c)
>(
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)
k•(
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)=(
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)
k+1也就是說,等式對n=k+1也成立
由①②知,a
n+c
n>2b
n對一切自然數(shù)n均成立
點評:本題主要考查數(shù)學歸納法證明不等式,等差數(shù)列、等比數(shù)列的性質及數(shù)學歸納法證明不等式的一般步驟.屬于綜合性試題有一定的難度.